Slutsky's theorem
/ SLOOT-skee /
In real statistics you rarely have a clean limit; you have a main quantity that is converging in distribution to something interesting (a normal, say) tangled together with a few nuisance quantities that are merely converging to constants — an estimated standard error, a sample variance. Slutsky's theorem tells you exactly when you may treat those nuisance terms as if they were already their constant limits, and combine everything.
Precisely: suppose X_n converges in distribution to X, and Y_n converges in PROBABILITY to a constant c. Then the combinations behave as you would hope: X_n + Y_n converges in distribution to X + c, the product X_n * Y_n converges in distribution to c*X, and X_n / Y_n converges in distribution to X/c (provided c is not 0). The key asymmetry is essential: one piece converges to a genuine random limit (in distribution), the OTHER must converge to a constant (in probability). You then just slot the constant in. This is what licenses replacing an unknown sigma by a consistent estimate without changing the limiting normal.
Why care, and where the fine print bites. Slutsky is the glue that makes confidence intervals and t-statistics work: (X-bar - mu)*sqrt(n) is asymptotically Normal(0, sigma^2), and dividing by a consistent estimate s_n of sigma still gives a standard normal limit, because s_n converges in probability to the constant sigma. The honest warning: the second sequence must converge to a CONSTANT, not to another random variable. If both pieces are genuinely random in the limit, you cannot just add their distributions — convergence in distribution does not respect addition of two random limits, precisely because it says nothing about whether they share outcomes.
To build a confidence interval for a mean, write T_n = (X-bar - mu)*sqrt(n) / s_n. The numerator converges in distribution to Normal(0, sigma^2); the denominator s_n converges in probability to the constant sigma. By Slutsky, T_n converges in distribution to Normal(0, 1) — so you may use the normal table even though sigma was estimated.
A nuisance term converging to a constant can be slotted in without disturbing the limiting distribution.
The second sequence must converge to a CONSTANT, not another random variable. You cannot generally add two convergence-in-distribution limits, since that mode says nothing about shared outcomes.